The Monster Integral

The problem looks difficult at the first sight. A little simplification makes the subsequent steps obvious.

From the numerator, collect the logarithmic terms first.

$\displaystyle \int_0^{\pi/4} x\frac{(1+x^2)+(x^2-1)\ln\left(\frac{1-x^2}{1+x^2}\right)}{(1-x^4)(1+x^2)}\exp\left(\frac{x^2-1}{x^2+1}\right)\,dx$

Rewrite $(1-x^4)=(1-x^2)(1+x^2)$ and divide the numerator by $(1+x^2)$ .

$\displaystyle \int_0^{\pi/4} \frac{x}{(1-x^2)(1+x^2)}\left(1+\frac{x^2-1}{x^2+1}\ln\left(\frac{1-x^2}{1+x^2}\right)\right)\exp\left(\frac{x^2-1}{x^2+1}\right)\,dx$

Use the substitution $\displaystyle \frac{x^2-1}{x^2+1}=t \Rightarrow \frac{4x}{(1+x^2)^2}\,dx=dt$ to get:

$\displaystyle \frac{1}{4}\int_{a}^{-1}\frac{e^t}{t}\left(1+t\ln(-t)\right)\,dt= \frac{1}{4}\int_{a}^{-1}e^t\left(\frac{1}{t}+\ln(-t)\right)\,dt$

where $\displaystyle a=\frac{\pi^2/16-1}{\pi^2/16+1}$

Since $\displaystyle \int e^x(f'(x)+f(x))\,dx=e^xf(x)+C$ , the above definite integral is:

$\displaystyle \frac{1}{4}\left(e^t \ln(-t) \right|_{a}^{-1}=-\frac{1}{4}e^a\ln(-a) \approx \boxed{0.284007}$

my solution is , to the some extent, same as pranav arora but i think it would have had given you a nightmare typing the latex of the question! $$ First combine the logs and factorise all factorisable terms and proceed as the routine method ! and remember when any repeated terms come it's most suitable to put it as y or something you like !